Using Cylindrical Coordinates to find volume of a solid How do I use Cylindrical Coordinates to find volume of a solid in the first Octant that is bounded by the cylinder $x^2 + y^2 = 2y$, the half cone $z = \sqrt{x^2 + y^2}$, and the $xy$-plane.
I have drawn the region of integration and obtained this:
$\int_0^2 \int_0^\sqrt{2y-y^2}\int_0^\sqrt{x^2 + y^2} dzdxdy$
Is this correct and from here were do I apply the cylindrical coordinates?
 A: I think I got it:
$\int_0^{\frac{\pi}{2}} \int_0^1\int_0^z r dzdrd\theta$
Can someone confirm if its correct.
A: The question makes only sense if the solid is the volume outside the half cone.
Otherwise the volume is infinite, because (almost) half of the cylinder which extends
to $+\infty$ in $z$ is in it. So we have to integrate over $x^2+y^2\ge z^2$.
For constant $z$ the section through the solid
is the area inside the circle $x^2+y^2-2y=0$ defined by the cylinder, which is inside the circle $x^2+(y-1)^2=1$,
but outside the circle $z^2=x^2+y^2$ defined by the cone. And since we're supposed to look only at one octant,
$z\ge 0, x\ge 0, y\ge 0$. The intersection of the 2 circles vanishes if the radius of
the circle at (0,0) is too large, which means $z\le 2$.
$$
V=\int_0^2 dz \int\int_{x^2+y^2\ge z^2, x^2+(y-1)^2\le 1} dx dy
$$
[Note that the lower limit in the partial integral $\int_0^{\sqrt{2y-y^2}}dx$ of the OP is wrong.]
In cylindrical coordinates $r^2=x^2+y^2$, $x=r\cos\theta, y=r\sin\theta$
we find the limits of $\theta=[\theta_0,\pi/2]$ where the circle $x^2+y^2=r^2$ intersects
the circle $x^2+(y-1)^2=1$:
$$
r^2\cos^2\theta_0+(r\sin\theta_0-1)^2=1;
$$
$$
r^2-2r\sin\theta_0+1=1;
$$
$$
\to \theta_0=\arcsin (r/2);
$$
So with the usual Jacobi Determinant factor $r$ in cylindrical coordinates
$$
V=\int_0^2 dz \int_z^2 r dr \int_{\arcsin(r/2)}^{\pi/2}d\theta
$$
$$
=\int_0^2 dz \int_z^2 r dr (\pi/2 - \arcsin(r/2))
$$
Here $\int r\arcsin (r/2) dr= (r^2/2-1)\arcsin(r/2)+r\sqrt{1-r^2/4}/2$, so
$$
V=\int_0^2 dz \left[\frac{\pi}{2} \frac{r^2}{2}-\left((\frac{r^2}{2}-1)\arcsin\frac{r}{2}+\frac{r\sqrt{1-r^2/4}}{2}\right)\right]\mid_{r=z}^2
$$
$$
=\int_0^2 dz \left[\frac{\pi}{2} - \frac{\pi}{2} \frac{z^2}{2}+\left((\frac{z^2}{2}-1)\arcsin\frac{z}{2}+\frac{z\sqrt{1-z^2/4}}{2}\right)\right]
$$
and this is evaluated via
$$
\int (\frac{z^2}{2}-1)\arcsin(z/2) dz = \frac{z^3}{6}\arcsin\frac{z}{2}-z\arcsin\frac{z}{2}+\frac{z^2-10}{9}\sqrt{1-z^2/4}
$$
and
$$
\int \frac{z}{2}\sqrt{1-z^2/4} dz= -\frac{1}{12}(4-z^2)^{3/2}.
$$
$$
V=16/9.
$$
